In triangle PQR,(4 hat{i}+3 hat{j}+6 hat{k}), | vedclass

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(D) Let the position vectors of vertices $P, Q, R$ be $\vec{p} = 4 \hat{i}+3 \hat{j}+6 \hat{k}$,$\vec{q} = 2 \hat{i}+2 \hat{j}+3 \hat{k}$,and $\vec{r}

Vedclass · Accepted Answer

$\frac{5}{2} \hat{i}+\frac{3}{2} \hat{j}+3 \hat{k}$

Vedclass · Answer

(D) Let the position vectors of vertices $P, Q, R$ be $\vec{p} = 4 \hat{i}+3 \hat{j}+6 \hat{k}$,$\vec{q} = 2 \hat{i}+2 \hat{j}+3 \hat{k}$,and $\vec{r} = 3 \hat{i}+\hat{j}+3 \hat{k}$.First,calculate the lengths of sides $PQ$ and $PR$:$PQ = |\vec{q} - \vec{p}| = |(2-4) \hat{i} + (2-3) \hat{j} + (3-6) \hat{k}| = |-2 \hat{i} - \hat{j} - 3 \hat{k}| = \sqrt{(-2)^2 + (-1)^2 + (-3)^2} = \sqrt{4+1+9} = \sqrt{14}$.$PR = |\vec{r} - \vec{p}| = |(3-4) \hat{i} + (1-3) \hat{j} + (3-6) \hat{k}| = |-\hat{i} - 2 \hat{j} - 3 \hat{k}| = \sqrt{(-1)^2 + (-2)^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$.Since $PQ = PR$,$	riangle PQR$ is an isosceles triangle with $PQ = PR$.In an isosceles triangle,the angle bisector of the vertex angle $(P)$ is also the median to the base $(QR)$.Therefore,the point of intersection $A$ of the angle bisector of $P$ with $QR$ is the midpoint of $QR$.$A = \frac{\vec{q} + \vec{r}}{2} = \frac{(2 \hat{i}+2 \hat{j}+3 \hat{k}) + (3 \hat{i}+\hat{j}+3 \hat{k})}{2} = \frac{5 \hat{i} + 3 \hat{j} + 6 \hat{k}}{2} = \frac{5}{2} \hat{i} + \frac{3}{2} \hat{j} + 3 \hat{k}$.

In $\triangle PQR$,$(4 \hat{i}+3 \hat{j}+6 \hat{k})$,$(2 \hat{i}+2 \hat{j}+3 \hat{k})$ and $(3 \hat{i}+\hat{j}+3 \hat{k})$ are the position vectors of the vertices $P, Q$ and $R$ respectively. Then the position vector of the point of intersection of the angle bisector of $P$ with $QR$ is

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